Chắc là \(q\left(x\right)=x^2-4????\)

\(f\left(2\right)=2^5+2^2+1=37\) ; \(f\left(-2\right)=-27\)

Do \(f\left(x\right)\) có 5 nghiệm nên f(x) có dạng:

\(f\left(x\right)=\left(x-x_1\right)\left(x-x_2\right)\left(x-x_3\right)\left(x-x_4\right)\left(x-x_5\right)\)

\(\Rightarrow f\left(2\right)=\left(2-x_1\right)\left(2-x_2\right)\left(2-x_3\right)\left(2-x_4\right)\left(2-x_5\right)=37\)

\(f\left(-2\right)=\left(-2-x_1\right)\left(-2-x_2\right)\left(-2-x_3\right)\left(-2-x_4\right)\left(-2-x_5\right)=-27\)

\(\Rightarrow\left(2+x_1\right)\left(2+x_2\right)\left(2+x_3\right)\left(2+x_4\right)\left(2+x_5\right)=27\)

\(A=\left(x_1^2-4\right)\left(x^2_2-4\right)\left(x_3^2-4\right)\left(x_4^2-4\right)\left(x^2_5-4\right)\)

\(A=-\left(2-x_1\right)\left(2-x_2\right)\left(2-x_3\right)\left(2-x_4\right)\left(2-x_5\right)\left(2+x_1\right)\left(2+x_2\right)\left(2+x_3\right)\left(2+x_4\right)\left(2+x_5\right)\)

\(A=-37.27=-999\)

 720 : ( x . 2 + x . 3 ) = 3.2
720 : ( x . 2 + x.3 ) = 6
( x .2 + x.3 )           = 720 : 6 
x.2+x.3 = 120
x . ( 2 + 3 ) = 120
x . 5 = 120
     x     = 120 : 5 
    x      = 24

Vì P(x) là đa thức bậc 4 và có 4 nghiệm x₁ , x₂ , x₃ , x₄ nên P(x) có thể viết thành : \(P\left(x\right)=\left(x-x_1\right)\left(x-x_2\right)\left(x-x_3\right)\left(x-x_4\right)\)

Xét :  \(Q\left(x\right)=x^2-4=\left(x-2\right)\left(x+2\right)=\left(2-x\right)\left(-2-x\right)\)

Ta có \(Q\left(x_1\right)=\left(2-x_1\right)\left(-2-x_1\right)\); \(Q\left(x_2\right)=\left(2-x_2\right)\left(-2-x_2\right)\); 

\(Q\left(x_3\right)=\left(2-x_3\right)\left(-2-x_3\right)\) ; \(Q\left(x_4\right)=\left(2-x_4\right)\left(-2-x_4\right)\)

Suy ra : \(T=Q\left(x_1\right).Q\left(x_2\right).Q\left(x_3\right).Q\left(x_4\right)\)

\(=\left[\left(2-x_1\right)\left(2-x_2\right)\left(2-x_3\right)\left(2-x_4\right)\right].\left[\left(-2-x_1\right)\left(-2-x_2\right)\left(-2-x_3\right)\left(-2-x_4\right)\right]\)

\(=P\left(2\right).P\left(-2\right)=-5.3=-15\)

Vậy T = -15

Vì P(x) có các nghiệm là x₁ , x₂ , x₃ , x₄ nên P(x) có thể viết được dưới dạng : \(P\left(x\right)=\left(x-x_1\right)\left(x-x_2\right)\left(x-x_3\right)\left(x-x_4\right)\)

Ta có : \(Q\left(x\right)=x^2-4=\left(x-2\right)\left(x+2\right)=\left(2-x\right)\left(-2-x\right)\)

Xét : \(Q\left(x_1\right)=\left(2-x_1\right)\left(-2-x_1\right)\) ; \(Q\left(x_2\right)=\left(2-x_2\right)\left(-2-x_2\right)\)

\(Q\left(x_3\right)=\left(2-x_3\right)\left(-2-x_3\right)\) ; \(Q\left(x_4\right)=\left(2-x_4\right)\left(-2-x_4\right)\)

Suy ra : 

\(T=Q\left(x_1\right).Q\left(x_2\right).Q\left(x_3\right).Q\left(x_4\right)=\left[\left(2-x_1\right)\left(2-x_2\right)\left(2-x_3\right)\left(2-x_4\right)\right]\left[\left(-2-x_1\right)\left(-2-x_2\right)\left(-2-x_3\right)\left(-2-x_4\right)\right]\)

\(=P\left(2\right).P\left(-2\right)\)

Bạn thay P(2) và P(-2) vào và tính nhé :)

Đa thức \(P\left(x\right)=x^4-5x^2-2x+3\)có bốn nghiệm là \(x_1;x_2;x_3;x_4\)nên P(x) có dạng \(\left(x-x_1\right)\left(x-x_2\right)\left(x-x_3\right)\left(x-x_4\right)\)(do P(x) là đa thức bậc bốn)

Ta có: \(Q\left(x\right)=x^2-3=\left(x+\sqrt{3}\right)\left(x-\sqrt{3}\right)\)

\(\Rightarrow T=Q\left(x_1\right).Q\left(x_2\right).Q\left(x_3\right).Q\left(x_4\right)\)

\(=\left[\left(x_1-\sqrt{3}\right)\left(x_2-\sqrt{3}\right)\left(x_3-\sqrt{3}\right)\left(x_4-\sqrt{3}\right)\right]\)

                   \(\left[\left(x_1+\sqrt{3}\right)\left(x_2+\sqrt{3}\right)\left(x_3+\sqrt{3}\right)\left(x_4+\sqrt{3}\right)\right]\)

\(=P\left(\sqrt{3}\right).P\left(-\sqrt{3}\right)=\left(-3-2\sqrt{3}\right)\left(-3+2\sqrt{3}\right)\)

\(=\left(3+2\sqrt{3}\right)\left(3-2\sqrt{3}\right)=9-12=-3\)

Vậy \(T=Q\left(x_1\right).Q\left(x_2\right).Q\left(x_3\right).Q\left(x_4\right)=-3\)

Mình nghĩ thế này bạn à:

PT1: \(x^2+2013x+2=0.\)Theo Hệ thức Vi-ét ta có: \(x_1+x_2=-2013\\ x_1.x_2=2\)

Tương tự với PT2 ta có:\(x_3+x_4=-2014\\ x_3.x_4=2\)

\(Q=\left[\left(x_1+x_3\right)\left(x_2-x_4\right)\right]\left[\left(x_2_{ }-x_3\right)\left(x_1+x_4\right)\right]\)

\(Q=\left(x_1.x_2+x_2.x_3-x_1.x_4-x_3.x_4\right)\left(x_1.x_2+x_2.x_4-x_1.x_3-x_3.x_4\right)\)

\(Q=\left(2+x_2.x_3-x_1.x_4-2\right)\left(2+x_2.x_4-x_1.x_3-2\right)\)

\(Q=\left(x_2.x_3-x_1.x_4\right)\left(x_2.x_4-x_1.x_3\right)\)

\(Q=x_2.x_3.x_4-x_3.x_1.x_2-x_4.x_1.x_2+x_1.x_3.x_4\)

\(Q=2x_2-2x_3-2x_4+2x_1\)

\(Q=2\left(x_1+x_2\right)-2\left(x_3+x_4\right)\)

\(Q=2.\left(-2013\right)-2.\left(-2014\right)\)

\(Q=2\)

Bài này hay quá. Chúc bạn học tốt nhé